What is the integral of tan^3(8x)sec(8x) ?

The integral of tan^3(8x)sec(8x) is -1/8 sec(8x)+(sec^3(8x))/(24)+C

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integral of tan^3(8x)sec(8x)

\int tan^{3} (8 x) sec (8 x) d x

- \frac{1}{8} sec (8 x) + \frac{sec ^{3} ( 8 x )}{24} + C

Solution steps

\int tan^{3} (8 x) sec (8 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (8 x)) tan (8 x) sec (8 x) d x

Apply u-substitution

: \int \frac{- 1 + u ^{2}}{8} d u

= \int \frac{- 1 + u ^{2}}{8} d u

Expand

\frac{- 1 + u ^{2}}{8} : - \frac{1}{8} + \frac{u ^{2}}{8}

= \int - \frac{1}{8} + \frac{u ^{2}}{8} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{8} d u + \int \frac{u ^{2}}{8} d u

\int \frac{1}{8} d u = \frac{1}{8} u

\int \frac{u ^{2}}{8} d u = \frac{u ^{3}}{24}

= - \frac{1}{8} u + \frac{u ^{3}}{24}

Substitute back

u = sec (8 x)

= - \frac{1}{8} sec (8 x) + \frac{sec ^{3} ( 8 x )}{24}

Add a constant to the solution

= - \frac{1}{8} sec (8 x) + \frac{sec ^{3} ( 8 x )}{24} + C

\int \frac{10}{x ^{4}} d x

\int \frac{6 x ^{2} + 2 x - 1}{3 x ^{2}} d x

\frac{d}{d x} (\frac{y}{x ^{2} + y ^{2} + 1})

f (x) = x^{2} e^{4 x}

\frac{d}{d x} (ln (x) sec (x))

What is the integral of tan^3(8x)sec(8x) ?
The integral of tan^3(8x)sec(8x) is -1/8 sec(8x)+(sec^3(8x))/(24)+C